Recent Questions
Q: Whenver I choose the enable Ajax feature and set the AjaxCount to 0, my menus don't pop out anymore. How do I get it to pop out?
A: That variable set the maximal number of submenus that will be loadedfrom server on your page.
If you don't know exact number of submenus you can set a big value forthis parameter.
If you set var dmAJAXCount=0; you cannot see you submenus.
To enable Ajax feature you should set, for example:
var dmAJAX=1;
var dmAJAXCount=100;
Q: How can I add a separator to the vertical menu using the dynamic functions?
It seems as though the only way to get a separator into the menu is by defining itbefore hand, but the website I'm building will require me to reload themenus dinamically, and I'm losing the ability to put separators on it.
A: You should add items using the following function:
function dm_ext_addItem (menuInd, submenuInd, iParams)
or
function dm_ext_addItemPos (menuInd, submenuInd, iParams, Pos)
Find more info:
http://deluxe-menu.com/functions-info.html
Set the following parameter:
var dm_writeAll=1;
So, you should write:
dm_ext_addItem(0, 4, ["|-", "", "", "", "", "", "", ])
Q: I need to work it with PHP/MySQL. I have found some information on it in your Support section, but have problems to understand and wonder if you have more information or a working sample of the loading bar with PHP/ MySQL.
A: Info about generating menu (menu items )from a database, please, see:
http://deluxe-menu.com/generate-menu-from-database-xml-php-asp-vb-support.html
Please, see the example of .php file.
The content of .php file depends on your database structure.
<?php
// The example for PHP/MySQL.
// MySQL database has the table "menuTable" that contains data for menu items.
// The table has the following fields:
// 1. "text" - item text
// 2. "link" - item link
// 3. "icon1" - item icon (normal state)
// 4. "icon2" - item icon (mouseover state)
function getMenuItems()
{
$jsItems = '';
// Select all records from table "menuTable"
$allItems = mysql_query('SELECT * FROM menuTable;');
// Extract items data from database and build Javascript code for menuItems
while ($itemData=mysql_fetch_array($allItems))
{
$jsItems .= '["'.$itemData['text'].'", "'.$itemData['link'].'", "'.$itemData['icon1'].'", "'.$itemData['icon2'].'"],';
}
// Return Javascript code
return $jsItems;
}
?>
<script>
var menuParam1 = value1;
var menuParam2 = value2;
var menuParam2 = value2;
...
var menuItems = [
<?php
// Write Javascript code for menu items
echo getMenuItems();
?>
];
</script>
Q: I used the drop menu creator to create a simple tab example, and I created a simple HTML file to display the tabs. It initially appears fine when the page first displays (e.g., the correct div is hidden), but when I *first* click on a tab I always get a javascript error deep in the .js code. Then I click around a bit between the tabs and get some display problems (no more javascript errors though), until finally the show/hide logic recovers and all looks ok.
The error I initially get is something like “tabs[…].id is not an object”. I have the IE7 debugger turned on, otherwise I might not ever see this error.
FYI, I tried your example menus on your web site and everything is fine! Maybe the .js files on your web site are not the same as those created by the generator?
A: You should set the following parameter in drop menu creator:
var bselectedSmItem=0;
